An oil drop of 12 excess electrons is held stationary under a constant electric field of `2.55xx10^(4) NC^(-1)` in Millikan's oil drop experiment. The density of the oil is 1.26 g `cm^(-3)`. Estimate the radius of the drop, `(g=9.81 ms^(-2), e=1.60xx10^(-19) C)`.

Explanation

Given `n=12, E=2.55xx10^(4) N//C`
`rho=1.26 g//cm^(3)=1.26xx10^(3) kg//m^(3)`
`e=1.6xx10^(-19) C, g=9.81 ms^(-2)`
As the oil drop is stationary,
Electostatic force = Gracitational force
`implies qE=mg`
`n eE=4/3 pi r^(3) rho g`
`r^(3)=(3 n eE)/(4 pi rho g)=(3xx12xx1.6xx10^(-19)xx2.55xx10^(4))/(4xx3.14xx1.26xx10^(3) xx9.8)`
`r^(3)=0.94xx10^(-18)`
`r=[0.94xx10^(-18)]^(1/3)=9.81xx10^(-7) m`
`:.` Radius of the drop `=9.81 xx10^(-7) m`.